Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561\, \mathring{A}$. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

• A. $1215\mathring{A}$
• B. $16405\mathring{A}$
• C. $2430\mathring{A}$
• D. $4687\mathring{A}$

Answer 1

The Correct Option is A

For hydrogen or hydrogen type atoms
$\frac{1}{\lambda}=RZ^2\bigg(\frac{1}{n^2_f}-\frac{1}{n^2_i}\bigg)$
In the transition from $n_i\rightarrow n_f$
$\therefore\lambda?\frac{1}{Z^2\bigg(\frac{1}{n^2_f}-\frac{1}{n^2_i}\bigg)}$
$\therefore\frac{\lambda_2}{\lambda_1}=\frac{Z^2_1\bigg(\frac{1}{n^2_f}-\frac{1}{n^2_i}\bigg)_1}{Z^2_2\bigg(\frac{1}{n^2_f}-\frac{1}{n^2_i}\bigg)_2},$
${\lambda_1}=\frac{\lambda_1Z^2_1\bigg(\frac{1}{n^2_f}-\frac{1}{n^2_i}\bigg)_1}{Z^2_2\bigg(\frac{1}{n^2_f}-\frac{1}{n^2_i}\bigg)_2},$
Substituting the values, we have
$=\frac{(6561\mathring{A})(1)^2\bigg(\frac{1}{2^2}-\frac{1}{3^2}\bigg)}{(2^2)\bigg(\frac{1}{2^2}-\frac{1}{4^2}\bigg)}=1215\mathring{A}$
$\therefore$ Correct option is (a).